Fair representation by independent sets

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1 Fair represetatio by idepedet sets Ro Aharoi Departmet of Mathematics, Techio, Haifa 32000, Israel Noga Alo Sackler School of Mathematics ad Blavatik School of Computer Sciece, Tel Aviv Uiversity, Tel Aviv, Israel Eli Berger Departmet of Mathematics, Haifa Uiversity, Haifa 31999, Israel Maria Chudovsky Departmet of Mathematics, Priceto Uiversity, Priceto, NJ 08544, USA Dai Kotlar Departmet of Computer Sciece, Tel-Hai College, Upper Galilee, Israel Marti Loebl Dept. of Applied Mathematics, Charles Uiversity, Praha, Malostraske.25, Praha 1 Czech Republic loebl@kam.mff.cui.cz Ra Ziv Departmet of Computer Sciece, Tel-Hai College, Upper Galilee, Israel raziv@telhai.ac.il Abstract Give a partitio of a set V ito pairwise disjoit sets V = (V 1, V 2,..., V m) ad a umber α 1, a subset S of V := V is said to represet V i α-fairly if S V i α V i, ad it is said to represet V α-fairly if it represets α-fairly all V is. We wish to represet early fairly (the meaig of early will traspire below) the sets V i with large α, by a set S of vertices that are idepedet i a give graph o V. We study the followig two cojectures: Cojecture 1. Suppose that the edges of K, are partitioed ito sets E 1, E 2,..., E m. The there 1, with strict iequality holdig for exists a perfect matchig F i K, satisfyig F E i all but at most oe value of i. Research supported by BSF grat o ad by the Discout Bak Chair at the Techio. Research supported by a USA-Israeli BSF grat, by a ISF grat, ad by the Israeli I-Core program. Research supported by BSF grat o ad by the Discout Bak Chair at the Techio. Research partially supported by BSF Grat No Gratefully ackowledges support by the Czech Sciece Foudatio GACR, cotract umber P G061, CE-ITI. Ei 1

2 ad: Cojecture 2. If P is a path ad V = V (F ) = V 1... V m is a partitio, the there exists a subset V m S of V that is idepedet i P, ad satisfies S, ad S V 2 i V i 1 for all i m. 2 These cojectures exted ad modify several well studied questios. We prove the existece of sets satisfyig either coditio of Cojecture 2 (ot ecessarily both), ad we prove Cojecture 1 for m = 2 ad m = 3. The proofs are i part topological, usig Sperer s lemma, the Borsuk-Ulam theorem ad a theorem of Schrijver o subgraphs of Keser s graph that are critical for the chromatic umber. Keywords: matchigs, idepedet sets, fair represetatio. 1 Itroductio: A cojecture of Ryser ad ramificatios Of the may directios from which the subject of this paper ca be approached, let us itroduce it through the les of a attractive cojecture - Ryser s Lati squares cojecture. Give a array A of symbols, a partial trasversal is a set of etries take from distict rows ad colums, ad cotaiig distict symbols. A partial trasversal of size is called simply a trasversal. Ryser s cojecture [27] is that if A is a Lati square, ad is odd, the A ecessarily has a trasversal. The oddess coditio is ideed ecessary - for every eve > 0 there exist Lati squares ot possessig a trasversal. A example is the additio table of Z : if a trasversal T existed for this Lati square, the the sum of its elemets, modulo, is k k = (+1) 2 (mod). O the other had, sice every row ad every colum is represeted i this sum, the sum is equal to i i + j j = ( + 1)(mod), ad for eve the two results do ot agree. Arsovski [14] proved a closely related cojecture, of Sevily, that every square submatrix (whether eve or odd) of the additio table of a odd order abelia group possesses a trasversal. Brualdi [17] ad Stei [32] cojectured that for ay, ay Lati square of order has a partial trasversal of order 1. Stei [32] observed that the same coclusio may follow from weaker coditios - the square does ot have to be Lati. Possibly it is eough to assume that the etries of the square are equally distributed betwee symbols. Put i the termiology of bipartite graphs, this reads: Cojecture 1.1. If the edge set of K, is partitioed ito sets E 1, E 2,..., E of size each, the there exists a almost perfect matchig i K, cosistig of oe edge from all but possibly oe E i. I this cojecture there is o distictio betwee eve ad odd. It is easy to costruct examples of squares satisfyig Stei s coditio, i which there is o full trasversal for eve as well as for odd. I matrix laguage, take a matrix M with m i,j = i for j <, ad m i, = i + 1(mod) (i particular, m, = 1). This cojecture belogs to a wider family of problems. Give a complex (closed dow hypergraph) C o a vertex set V, ad a partitio V of V ito sets V 1, V 2,..., V m, we may ask for the largest umber α for which there exists a subset S of V belogig to C, ad satisfyig S V i α V i for all i, or, as i Stei s cojecture, for all values of i but oe, or for all i but a fixed umber. A tight boud for α ca be obtaied i the case of matroids. For a give complex C, let β(c) be the miimal umber of edges (simplices) of C whose uio is V (C). A result followig directly from Edmods matroid itersectio theorem is: Theorem 1.2. If C is a matroid the for every partitio V of V (C) there exists a set S C satisfyig S V i for all i. Vi β(c) We shall maily be iterested i the case that C is the complex of idepedet sets of a graph G, deoted by I(G). The followig theorem of Haxell [20] pipoits the right value of α i this case: 2

3 Theorem 1.3. If V = (V 1, V 2,..., V m ) is a partitio of the vertex set of a graph G, ad if V i 2 (G) for all i m, the there exists a set S idepedet i G, itersectig all V i s. This was a improvemet over earlier results of Alo, who proved the same with 25 (G) [10] ad the with 2e (G) [11]. Corollary 1.4. If the vertex set V of a graph G is partitioed ito sets V 1, V 2,..., V m the there exists a idepedet subset S of V, satisfyig S V i for every i m. Vi 2 (G) Proof. For each i m pack Vi 2 (G) disjoit sets of size 2 (G) (call them V j i ) i each V i. By Theorem 1.3 there exists a idepedet set S meetig all V j i, ad this is the set desired i the theorem. So, for C = I(G), the complex of idepedet sets i G, the magic umber is α = 1 2. This is sharp, as show i [35, 21, 33]. Note that β(i(g)) = 1 χ(g), with strict iequality holdig i geeral, meaig that I(G) does ot behave so well with respect to represetatio as matroids do. The secret coectig this result to Stei s cojecture ad to Cojecture 1.1 is that for lie graphs much better bouds ca be obtaied. Theorem 1.5. If H is a graph ad G = L(H) (the lie graph of H) the there exists a idepedet set S such that S V i for every i m. Vi (G)+2 This follows from a result, proved i [3], that if G is a lie graph of a graph the the topological coectivity of I(G), deoted by η(i(g)), satisfies η(i(g)) V (G) + 2 (1) I [7] a geeralizatio of (1) was proved for hypergraphs: If H is a hypergraph ad G = L(H) the η(i(g)) V max e H v e deg H(v). (2) To see why (2) geeralizes (1) ote that if H is r-uiform ad liear (o two edges meet at more tha oe vertex), the (G) = max e H v e deg H(v) r, which together with (2) etails that η(i(g)) V (G)+r. The coectivity η(c) of a complex C is the miimal dimesio of a hole i C, so for example if C has a o-empty vertex set but is ot path coected the η(c) = 1, sice there is a hole of dimesio 1, cosistig of two poits that caot be joied by a path (it is this o existig path that is the hole of dimesio 1). The way from (1) to Theorem 1.5 goes through a topological versio of Hall s theorem, proved i [6]. I a eve stroger versio of Cojecture 1.1, the sigle E i that is ot represeted ca be arbitrarily chose. To put this formally, we shall use the followig termiology: Defiitio 1.6. A raibow set of a collectio S 1, S 2,..., S m is a choice fuctio from these sets. If S i are sets of graph edges ad the chose edges form a matchig, the the raibow set is called a raibow matchig. Cojecture 1.7. Ay family E 1, E 2,..., E 1 of disjoit subsets of size of E(K, ) has a raibow matchig. 3

4 The three coditios - that the sets E i are disjoit, that they are subsets of E(K, ), ad that their umber is 1, seem a bit artificial. It is eticig to make the followig cojecture, that etails the case i which the sets E i are matchigs: If E 1, E 2,..., E m are sets of edges i a bipartite graph, ad E i > ( i m E i) (where ( i m E i) is the maximal degree of a vertex i the multigraph i m E i) the there exists a raibow matchig. Ufortuately, this cojecture is false, as show by the followig example: Example 1.8. [21, 35] Take three vertex disjoit copies of C 4, say A 1, A 2, A 3. Number the edges of A i cyclically as a j i (j = ). Let E 1 = {a 1 1, a 3 1, a 1 3}, E 2 = {a 2 1, a 4 1, a 3 3}, E 3 = {a 1 2, a 3 2, a 2 3} ad E 4 = {a 2 2, a 4 2, a 4 3}. The ( i m E i) = 2, E i = 3 ad there is o raibow matchig. I [4] the followig was suggested: Cojecture 1.9. If E 1, E 2,..., E m are sets of edges i a bipartite graph, ad E i > ( i m E i) + 1 the there exists a raibow matchig. Re-phrased, this cojecture reads: If H is a bipartite multigraph, G = L(H) ad V i V (G) satisfy V i (H) + 2 for all i, the there exists a idepedet set i G (amely a matchig i H) meetig all V i s. We do ot kow of other examples, beyod Example 1.8, i which V i (H) + 1 does ot suffice. The cojecture is false if the sets V i are allowed to be multisets - we omit the details of the example showig this. Cojecture 1.9 would yield: Cojecture If the edge set of a graph H is partitioed ito sets F 1,..., F m the there exists a matchig M satisfyig M F i for all i m If true, Cojecture 1.7 would imply: Fi (H)+2 Cojecture Suppose that E(K, ) is partitioed ito sets E 1, E 2,..., E m. The there exists a 1, with strict iequality holdig for all but oe perfect matchig F i K, satisfyig F E i value of i. This ca be stregtheed to: Ej Ei Cojecture I Cojecture 1.11 it is possible to choose the idex for which the strict iequality does ot occur. Namely, for every j m there exists a perfect matchig F i K, satisfyig F E i Ei for all i j, ad F E j 1. We shall prove: Theorem Cojecture 1.12 is true for m = 2, 3, for all. For m = 2 we shall also characterize the cases where ( 1) is eeded for oe of the idices. Lie graphs are oe family of graphs where Theorem 1.2 ca be improved upo. Aother is paths. Cojecture Give a path P ad a partitio of V (P ) ito sets V 1,..., V m there exists a idepedet set S such that S V i Vi 2 1 for all i, with strict iequality holdig for all but at most m 2 sets V i. While the full cojecture is ope, we shall prove the existece of a idepedet set satisfyig either oe of the two coditios. 4

5 Theorem Give a partitio of the vertex set of a path ito sets V 1,..., V m there exists a idepedet set S ad itegers b i, i m, such that i m b i m 1 2 ad S V i Vi 2 b i for all i. The secod result we shall prove applies also to cycles. Of course, Theorem 1.15 ca be proved also for cycles, up to oe vertex: Theorem Give a partitio of the vertex set of a cycle ito sets V 1,..., V m there exists a idepedet set S such that S V i Vi 2 1 for all i. The proofs of three of our mai results - that of Theorem 1.13 for the case m = 3, ad those of Theorems 1.15 ad 1.16, are topological. The first uses Sperer s lemma, ad the secod uses the Borsuk-Ulam theorem. The proof of Theorem 1.16 uses a theorem of Schrijver, stregtheig a famous theorem of Lovász o the chromatic umber of Keser graphs. This meas that it, too, uses the Borsuk-Ulam theorem, sice the Lovász-Schrijver proof uses the latter. We refer the reader to Matousek s book [25] for backgroud o topological methods i combiatorics, i particular the use of the Borsuk-Ulam theorem. 2 Fair represetatio by idepedet sets i paths: a Borsuk- Ulam approach I this sectio we prove Theorem Followig a idea from the proof of the ecklace theorem [9], we shall use the Borsuk-Ulam theorem. I the ecklace problem two thieves wat to divide a ecklace with m types of beads, each recurrig i a eve umber of beads, so that the beads of every type are evely split betwee the two. The theorem is that the thieves ca achieve this goal usig at most m cuts of the ecklace. I our case, the two thieves are the sets of odd ad eve poits, respectively. But rather tha use the theorem as a black box, we have to adapt ideas from its proof to the preset situatio. Proof of Theorem 1.15 Let v 1,..., v be the vertices of P, ordered alog the path. Our aim is to form a idepedet set meetig each of the sets V i partitioig V i approximately V i /2 vertices. I order to use the Borsuk- Ulam theorem, we first make the problem cotiuous, by replacig each v p by the characteristic fuctio of the pth of itervals of legth 1 i [0, 1], amely by ṽ p = χ [ p 1, p bead. p 1 ]. We call the iterval [, p ] a As usual, S m 1 deotes the set of poits x = (x 1,..., x m ) R m satisfyig i m x2 i = 1. Give such a poit x, let z k = j k x2 j (z 0 = 0). Let g be the characteristic fuctio of the odd beads o the path. Explicitly, for every 1 p odd let g(y) = 1 for all p 1 y p p 1 ad for every 1 p eve let g(y) = 0 for all y p. Let h be the characteristic fuctio of the eve beads, amely h(y) = 1 g(y). For every i m let χ i be the sum of all characteristic fuctios of beads belogig to V i. That is, for every 0 p 1 ad all p 1 y p let χ i(y) = 1 if v p V i ad χ i (y) = 0 otherwise. For every i m defie a fuctio f i : S m 1 R m 1 by: f i (x 1,..., x m ) = 1 k m zk z k 1 (g(y) h(y))χ i (y)sig(x k )dy Here, as usual, sig(x) = 0 if x = 0 sig(x) = 1 if x > 0 ad sig(x) = 1 if x < 0. The fuctios f i are cotiuous, because at the poits of discotiuity of the sig fuctio the itervals i the itegrals are of zero legth. The sig term guaratees that f i ( x) = f i ( x). Hece, by the Borsuk-Ulam theorem there exists a poit w = (w 1,..., w m ) S m 1 such that f i ( w) = 0 for all i [m], 5

6 where z k = j k w2 j. For y [0, 1] such that y [z k 1, z k ) defie P OS(y) = 1 if w k 0 ad P OS(y) = 0 otherwise. Let NEG(y) = 1 P OS(y). Let v(y) = P OS(y)g(y)+NEG(y)h(y). So, v chooses the odd beads (=vertices) from itervals o which w is positive, ad the eve beads from itervals o which w is egative. This choice does ot ecessarily costitute a idepedet set of vertices - the desired idepedet set I will be a subset of this choice. Fix ow i [m]. The fact that f i ( w) = 0 meas that 1 y=0 Shufflig terms this gives: 1 y=0 χ i (y)p OS(y)[g(y) h(y)]dy = χ i (y)[p OS(y)g(y) + NEG(y)h(y)]dy = 1 y=0 1 y=0 χ i (y)neg(y)[g(y) h(y)]dy χ i (y)[p OS(y)h(y) + NEG(y)g(y)]dy O the left had side there is the measure of the set of poits of V i chose by v, ad o the roght had side the measure of the set of poits of V i ot chose by v. So, 1 χ i (y)[p OS(y)g(y) + NEG(y)h(y)]dy = 1 χ i (y)dy y=0 2 y=0. Let F be the set of beads cotaiig a poit z k for some k. Let I 1 be the set of those odd beads that meet itervals (z k 1, z k ] i which w k 0 ad eve beads that meet itervals (z k 1, z k ] i which w k < 0. Let I 2 be the set of those eve beads that meet itervals (z k 1, z k ] i which w k 0 ad odd beads that meet itervals (z k 1, z k ] i which w k < 0. Note that the two may overlap at beads cotaiig poits z k. We shall choose as I either I 1 \ F or I 2 \ F. Let us show that for oe of these choices I 2 m 1 2. Note that i passig from the beads (livig i [0, 1]) to the sets I j (livig i []) quatities should be multiplied by Deote by αj k the losses icurred by the beads belogig to I j upo reoucig the beads cotaiig z k. The, clearly, α1 k + α2 k = m 1. With the -factor ivolved i passig to the sizes of I j, it follows that I 1 \ F + I 2 \ F (m 1). Takig I as the larger of I 1 \ F, I 2 \ F, I satisfies the coditios of the theorem. Remark We do ot kow whether it is always possible to divide the losses evely betwee the various V i s. I particular, we do ot kow whether i fact all b i s i the theorem ca assume oly the values 0 or The iequality i m b i m 1 2 caot be improved. Namely, there are examples i which the miimum of the sum i m b i i the theorem is m 1 2. To see this, let m = 2k + 1, ad let each V i be of size 2k. Cosider a sequece of legth 2k (2k + 1), i which the (i 1)m + 2j 1-th elemet belogs to V i (i = 1,..., 2k, j = 1,..., k + 1), ad the rest of the elemets are chose i ay way so as to satisfy the coditio V i = 2k. For example, if k = 2 the the sequece is of the form: where the s ca be filled i ay way that satisfies V i = 4 (amely, four of them are replaced by the symbol 5, ad oe is replaced by i for each symbol i = 1, 2, 3, 4), ad the dashes are just 6

7 for ease of referece to the four stretches. If S is a idepedet set i the path the, removig occurreces of symbol i so as to have at most k occurreces of i (for example, i the first stretch of the example above there is o poit i choosig all three 1s), we may assume that S cotais o more tha k elemets from each stretch, ad thus S 2k k, which is m 1 2 short of half the legth of the path. 3. It may be of iterest to fid the best bouds as a fuctio of the sizes of the sets V i ad their umber. Note that i the example above the size of the sets is almost equal to their umber. As oe example, if If all V i s are of size 2, the the iequality ca be improved to: i m b i m 3. To see this, look at the multigraph obtaied by addig to P the pairs formig the sets V i as edges. I the resultig graph the maximum degree is 3, ad hece by Brooks theorem it is 3-colorable (i fact, the theorem eeds ot be ivoked: the average degree i every iduced subgraph is less tha 3). Thus there is a idepedet set of size at least 3, which represets all V is apart from at most m 3 of them. 3 Fair represetatio by idepedet sets i cycles: usig a theorem of Schrijver I this sectio we shall prove Theorem The proof uses a result of Schrijver [28], a stregtheig of a theorem of Lovász: Theorem 3.1 (Schrijver [28]). For itegers k, satisfyig > 2k let K = K(, k) deote the graph whose vertices are all idepedet sets of size k i a cycle C of legth, where two such vertices are adjacet iff the correspodig sets are disjoit. The the chromatic umber of K is 2k + 2. I a more straightforward formulatio, this reads: Theorem 3.2. The family I(, k) of idepedet sets of size k i the cycle C caot be partitioed ito fewer tha 2k + 2 itersectig families. We start with a simple case, i which all V i but oe are odd: Theorem 3.3. Let m, r 1, r 2,..., r m be positive itegers, ad put = m i=1 (2r i + 1) 1. Let G = (V, E) be a cycle of legth, ad let V = V 1 V 2... V m be a partitio of its set of vertices ito m pairwise disjoit sets, where V i = 2r i + 1 for all 1 i < m ad V m = 2r m. The there is a idepedet set S of G satisfyig S = m i=1 r i ad S V i = r i for all 1 i m. Proof of Theorem 3.3: Put k = m i=1 r i ad ote that 2k + 2 = m + 1 > m. Assume, for cotradictio, that there is a partitio with parts V i of the set of vertices V of G as i the theorem, with o S I(, k) satisfyig the assertio of the theorem. The for every S I(, k) there is at least oe idex i for which S V i r i + 1. Ideed, otherwise S V i r i for all i ad hece S V i = r i for all i, cotradictig the assumptio. Let F i be the family of sets S I(, k) for which S V i r i + 1. Clearly, F i is itersectig (i fact, itersectig withi V i ), cotradictig the coclusio of Theorem 3.2. Corollary 3.4. If V = V 1 V 2 V m is a partitio of the vertex set of a cycle C ito m pairwise disjoit sets, the the followig hold: (i) If V i is eve for some idex i the there is a idepedet set S i of C so that S i V i = V i /2, for every j so that V j is odd, S i V j = ( V j 1)/2 ad for every j i so that V i is eve S V j = V j /2 1. (ii) If V i is odd for all i the for ay vertex v of C there is a idepedet set S of C so that v S ad S V i = ( V i 1)/2 for all i. 7

8 Proof of Corollary 3.4: Part (i) i case all sets V j besides V i are of odd sizes is exactly the assertio of Theorem 3.3. If there are additioal idices j i for which V j is eve, choose a arbitrary vertex from each of them ad cotract a edge icidet with it. The result follows by applyig the theorem to the shorter cycle obtaied. Part (ii) is proved i the same way, cotractig a edge icidet with v. 4 More applicatios of Schrijver s theorem ad its extesios 4.1 Hypergraph versios The results above ca be exteded by applyig kow hypergraph variats of Theorem 3.1. For itegers s 2, let C s 1 deote the (s 1)-th power of a cycle of legth, that is, the graph obtaied from a cycle of legth by coectig every two vertices whose distace i the cycle is at most s 1. Thus if s = 2 this is simply the cycle of legth whereas if 2s 1 this is a complete graph o vertices. For itegers, k, s satisfyig > ks, let K(, k, s) deote the followig s-uiform hypergraph. The vertices are all idepedet sets of size k i C s 1, ad a collectio V 1, V 2,..., V s of such vertices forms a edge iff the sets V i are pairwise disjoit. Note that for s = 2, K(, k, 2) is exactly the graph K(, k) cosidered i Theorem 3.1. The followig cojecture appears i [12]. Cojecture 4.1. For > ks, the chromatic umber of K(, k, s) is ks+s s 1. This is proved i [12] if s is ay power of 2. Usig this fact we ca prove the followig. Theorem 4.2. Let s 2 be a power of 2, let m ad r 1, r 2,..., r m be itegers, ad put = s m i=1 r i + (s 1)(m 1). Let V 1, V 2,..., V m be a partitio of the vertex set of C s 1 ito m pairwise disjoit sets, where V i = sr i + s 1 for all 1 i < m, ad V m = sr m. The there exists a idepedet set S i C s 1 satisfyig S V i = r i for all 1 i m. Proof: Put k = m i=1 r i ad ote that the chromatic umber of K(, k, s) is ( ks+s)/(s 1) > m. Assume, for cotradictio, that there is a partitio of the vertex set of C s 1 with parts V i as i the theorem, with o idepedet set of C s 1 of size k = m i=1 r i satisfyig the assertio of the theorem. I this case, for ay such idepedet set S there is at least oe idex i so that S V i r i + 1. We ca thus defie a colorig f of the idepedet sets of size k of C s 1 by lettig f(s) be the smallest i so that S V i r i + 1. Sice the chromatic umber of K(, k, s) exceeds m, there are s pairwise disjoit sets S 1, S 2,..., S s ad a idex i so that S j V i r i + 1 for all 1 j s. But this implies that V i sr i + s, cotradictig the assumptio o the size of the set V i, ad completig the proof. Just as i the previous sectio, this implies the followig. Corollary 4.3. Let s > 1 be a power of 2. Let V 1, V 2,..., V m be a partitio of the set of vertices of C s 1 where = m i=1 V i, ito pairwise disjoit sets. The there is a idepedet set S i C s 1 so that Vi s + 1 S V i = s for all 1 i < m, ad Vi S V m =. s The proof is by cotractig edges, reducig each set V i to oe of size s ad reducig V m to a set of size s graph. Vm s Vi s+1 s, +s 1 for 1 i < m,. The result follows by applyig Theorem 4.2 to this cotracted 8

9 4.2 The Du-Hsu-Wag cojecture Du, Hsu ad Wag [18] cojectured that if a graph o 3 vertices is the edge disjoit uio of a Hamilto cycle of legth 3 ad vertex disjoit triagles the its idepedece umber is. Erdős cojectured that i fact ay such graph is 3 colorable. Usig the algebraic approach i [13], Fleischer ad Stiebitz [29] proved this cojecture i a stroger form - ay such graph is i fact 3-choosable. The origial cojecture, i a slightly stroger form, ca be derived from Theorem 3.3: omit ay vertex ad apply the theorem with r i = 1 for all i. So, for every vertex v there exists a represetig set as desired i the cojecture omittig v. The derivatio of the statemet of Theorem 3.3 from the result of Schrijver i [28] actually supplies a quick proof of the followig: Theorem 4.4. Let C 3 = (V, E) be cycle of legth 3 ad let V = A 1 A 2... A be a partitio of its vertex set ito pairwise disjoit sets, each of size 3. The there exist two disjoit idepedet sets i the cycle, each cotaiig oe poit from each A i. Proof. Defie a colorig of the idepedet sets of size i C 3 as follows. If S is such a idepedet set ad there is a idex i so that S A i 2, color S by the smallest such i. Otherwise, color S by the color + 1. By [28] there are two disjoit idepedet sets S 1, S 2 with the same color. This color caot be ay i, sice if this is the case the (S 1 S 2 ) A i = S 1 A i + S 2 A i = 4 > 3 = A i, which is impossible. Thus S 1 ad S 2 are both colored + 1, meaig that each of them cotais exactly oe elemet of each A i. The Fleischer-Stiebitz theorem implies that the represetig set i the HDW cojecture ca be required to cotai ay give vertex. This ca also be deduced from the topological versio of Hall s Theorem proved i [6] (for this derivatio see e.g [2]). The latter shows also that the cycle of legth 3 ca be replaced by the uio of cycles, totallig 3 vertices, oe beig of legth 1 mod 3. Simple examples show that the Fleischer-Stiebitz theorem o 3-colorability does ot apply to this settig. Note that oe of the above proofs supplies a efficiet algorithm for fidig the desired idepedet set. 5 Fair represetatio by matchigs i K,, the case of two parts The case m = 2 of Cojecture 1.11 is easy. Here is its statemet i this case: Theorem 5.1. If F is a subset of E(K, ), the there exists a perfect matchig N such that N F 1 ad N \ F 1. F E(G)\F The fact that it is possible to reach ay permutatio from ay other by a sequece of traspositios meas that it is possible to reach every perfect matchig i K, from ay other by a sequece of exchages, i each step replacig two edges of the perfect matchig by two other edges. If such a sequece starts with surplus of F edges ad eds with shortage of F edges, the at the stage i which the trasitio from surplus to shortage occurs the coditio is satisfied. Thus, i this case the iterestig questio is i which cases is the ( 1) term ecessary. That this term is sometimes ecessary is show, for example, by the case of = 2 ad F beig a perfect matchig. Aother example - = 6 ad F = [3] [3] {4, 5, 6} {4, 5, 6}: it is easy to see that there is o perfect matchig cotaiig precisely 3 edges from F, as required i the cojecture. The appropriate coditio is give by the followig cocept: Defiitio 5.2. A subset F of E(K, ) is said to be rigid if there exist subsets K ad L of [] such that F = K L ([] \ K) ([] \ L). 9

10 The rigidity i questio is with respect to F -parity of perfect matchigs: Theorem 5.3. [8] A subset F of E(K, ) is rigid if ad oly if P F has the same parity for all perfect matchigs P i K,. This characterizatio shows that whe F is rigid, it is ot always possible to drop the mius 1 term i Theorem 5.1. We shall show: Theorem 5.4. If a subset F of E(K, ) is ot rigid, or if F, the there exists a perfect matchig N such that N F ad N \ F. F This will clearly follow from: E(G)\F Theorem 5.5. If a subset F of E(K, ) is ot rigid the for every iteger c such that ν(e(g)\f ) c ν(f ) there exists a perfect matchig N satisfyig N F = c. This theorem, i tur, easily follows from: Theorem 5.6. Let G = K, ad let a < c < b be three itegers. Suppose that F E(G) is ot rigid. If there exists a perfect matchig P a such that P a F = a ad a perfect matchig P b such that P b F = b, the there exists a perfect matchig P c satisfyig P c F = c. Proof. We use the matrix laguage of the origial Ryser cojecture. Let M be the matrix i which m(i, j) = 1 if (i, j) F ad m(i, j) = 0 if (i, j) F. A perfect matchig i G correspods to a geeralized diagoal (below, g.d ) i M, amely a set of etries belogig to distict rows ad colums. A g.d will be called a k-g.d if exactly k of its etries are 1. By assumptio there exist a a-g.d T a ad a b-g.d T b. Assume, for cotradictio, that there is o c-g.d. The case = 2 is trivial, ad hece, reversig the roles of 0s ad 1s if ecessary, we may assume that c > 1. Sice a g.d correspods to a permutatio i S, ad sice every permutatio ca be obtaied from ay other permutatio by a sequece of traspositios, there exists a sequece of g.d s T a = T 1, T 2,..., T k = T b, where each pair T i ad T i+1, i = 1,..., k 1, differ i two etries. By the egatio assumptio there exists i such that T := T i+1 is a c + 1-g.d ad T := T i is a (c 1)-g.d. Without loss of geerality we may assume that T lies alog the mai diagoal ad that its first c + 1 etries are 1. Let I = [c + 1], J = [] \ I ad let A = M[I I], B = M[I J], C = M[J I], D = M[J J] (we are usig here a commo otatio - M[I J] deotes the submatrix of M iduced by the row set I ad colum set J). We may assume that the g.d T is obtaied from T by replacig the etries (c, c) ad (c + 1, c + 1) by (c + 1, c) ad (c, c + 1) (Figure 1). Figure 1 10

11 Claim 1. The matrices A ad D are symmetric. Proof of Claim 1. To prove that A is symmetric, assume, for cotradictio, that there exist i 1 i 2 I such that m i1,i 2 m i2,i 1. The, we ca replace the etries (i 1, i 1 ) ad (i 2, i 2 ) i T by (i 1, i 2 ) ad (i 2, i 1 ) to obtai a c-g.d. The proof for D is similar, applyig the replacemet i this case to T. Claim 2. If i I ad j J the m i,j m j,i. Proof of Claim 2. Case I: m i,j = m j,i = 0. Replacig (i, i) ad (j, j) i T by (i, j) ad (j, i) results i a c-g.d. Case II: m i,j = m j,i = 1. Subcase II 1 : i {c, c + 1}. Replacig i T the etries (i, i) ad (j, j) by (i, j) ad (j, i) results i a c-g.d. Subcase II 2 : i {c, c+1}. Without loss of geerality we may assume i = c+1 ad j = c+2 (Figure 2). If m k,m = m m,k = 0 for some 1 k < m c the replacig i T the etries (k, k), (m, m), (c + 1, c + 1) ad (c + 2, c + 2) by (k, m), (m, k), (c + 1, c + 2) ad (c + 2, c + 1) results i a c-g.d. (Figure 2. I all figures the removed etries are struck out by ad the added etries are circled). Thus we may assume that m k,m = m m,k = 1 for all k, m c. Figure 2 We ow cosider three sub-subcases: (i) m c,c+2 = 0, m c+2,c = 1. I this case we may replace the etries (c, c), (c + 1, c + 1) ad (c + 2, c + 2) i T by (c, c + 2), (c + 1, c) ad (c + 2, c + 1) ad obtai a c-g.d (Figure 3(a)). (ii) m c,c+2 = 1, m c+2,c = 0. Replace the same etries as i Case (i) by (c, c + 1), (c + 1, c + 2) ad (c + 2, c), agai obtaiig a c-g.d (Figure 3(b)). (iii) m c,c+2 = m c+2,c = 1. If m c 1,c+1 = 0 the, rememberig that m(c 1, c 1) = 1, we ca replace (c 1, c 1), (c, c), (c + 1, c + 1) ad (c + 2, c + 2) i T by (c 1, c + 1), (c, c) + 2), (c + 1, c 1), (c + 2, c) ad obtai a c-g.d (Figure 4(a)). If m c 1,c+1 = 1, we ca replace (c 1, c 1), (c, c) ad (c + 1, c + 1) i T by (c 1, c + 1), (c, c 1) ad (c + 1, c) ad obtai a c-g.d (Figure 4(b).). This proves Claim 2. For a matrix K idexed by ay set of idices J deote by K (i) the ith row of K, ad by K (j) the jth colum of K. Claim 3. For ay j J, the submatrix A is the additio table modulo 2 of the row C (j) ad the colum B (j). (Figure 5). 11

12 (a) (b) Figure 3 (a) (b) Figure 4 Figure 5 Proof of Claim 3. We eed to show that for ay i 1, i 2 I ad j J we have m i1,i 2 = m j,i2 + m i1,j (mod 2). We may assume that i 1 i 2 sice the case i 1 = i 2 follows from Claim 2 ad the fact that A has 1 s i the mai diagoal. Let x = m j,i2 C (j) ad y = m i1,j B (j). We cosider three cases: (i) x y, (ii) x = y = 0, ad (iii) x = y = 1. 12

13 (i) Assume, for cotradictio, that m i1,i 2 = 0. The, by Claim 1, m i2,i 1 = 0 ad we ca replace (i 1, i 1 ), (i 2, i 2 ) ad (j, j) i T by (i 2, i 1 ), (i 1, j) ad (j, i 2 ) ad obtai a c-g.d (Figure 6(a)). (ii) Assume, for cotradictio, that m i1,i 2 = 1. We perform the same exchage as i Case (i) ad, agai, obtai a c-g.d (Figure 6(b)). (iii) By Claim 2, we have m i2,j = m j,i1 = 0. Assume, for cotradictio, that m i1,i 2 = 1. We replace (i 1, i 1 ), (i 2, i 2 ) ad (j, j) i T by (i 1, i 2 ), (i 2, j) ad (j, i 1 ) ad obtai a c-g.d (Figure 6(c)). This proves Claim 3. (a) (b) (c) Figure 6 We say that two (0,1)-vectors u ad v of the same legth are complemetary (deoted u v) if their sum is the vector (1, 1,..., 1). By Claim 3, for every i 1, i 2 I, if for some, or equivaletly ay, j J, it is true that m i1,j = m i2,j the the two rows A (i1), A (i2) are idetical, ad if m i1,j m i2,j the these two rows are complemetary. Furthermore - the rows M i1, M i2 are idetical or complemetary. We summarize this i: Claim 4. Ay two rows i M[I []] are either idetical or complemetary. Next we show that the property i Claim 4 holds for ay two rows i M. For x, y {0, 1} we defie the operatio x y = x + y + 1 (mod 2) (Figure 7). Claim 5. The submatrix D is the -table betwee the colum C (i) ad the row B (i), for ay i I. Figure 7 13

14 Proof of Claim 5. We first cosider i such that 1 i c 1 (we assumed c > 1). Let j 1, j 2 J. We may assume that j 1 j 2 sice the case j 1 = j 2 follows from Claim 2 ad the fact that D has 0 s i the diagoal. Let x = m j2,i ad y = m i,j1. We cosider three cases: (i) x = y = 0, (ii) x = y = 1, ad (iii) x y. (i) Assume, for cotradictio, that m j2,j 1 = 0. By Claim 1, m j1,j 2 = 0, ad we ca replace (i, i), (j 1, j 1 ) ad (j 2, j 2 ) i T by (i, j 1 ), (j 1, j 2 ) ad (j 2, i) ad obtai a c-g.d (Figure 8(a)). (ii) By Claim 2, m j1,i = m i,j2 = 0, ad we ca replace the same etries as i Case 1 by (i, j 2 ), (j 1, i) ad (j 2, j 1 ) ad obtai a c-g.d (Figure 8(b)). (iii) Here is where we eed the assumptio i c 1. We perform the same replacemet as i Case 1, but this time o the g.d T, ad obtai a c-g.d (Figure 8(c). Recall that T is a (c 1)-g.d). (a) (b) (c) Figure 8 It remais to prove the claim for i = c, c + 1. It follows from Claim 4 that ay two rows of B are either idetical or complemetary. Thus, by Claim 2, ay two colums of C are either idetical or complemetary. If there exists j < c such that B (c) = B (j), the C (c) = C (j). Sice D is the -table betwee C (j) ad B (j), it is also the -table betwee C (c) ad B (c). If all j < c satisfy B (c) B (j), the for ay such j, we have C (c) = B(j) T ad C(j) = B(c) T by Claim 2. Sice is commutative we agai have that D is the -table betwee C (c) ad B (c). A similar argumet holds for i = c + 1. Claim 6. Ay two rows of M are either idetical or complemetary. Proof of Claim 6. The fact that ay two rows i M[J []] are either idetical or complemetary follows i the same maer as Claim 4. Now, assume i I, j J. We wat to show that M (i) is either idetical or complemetary to M (j). From Claim 3 we kow that A (i) is either idetical or complemetary to C (j) ad from Claim 5 we have that B (i) is either idetical or complemetary to D (j). We eed to show that A (i) is idetical to C (j) if ad oly if B (i) is idetical to D (j). Note that m ii = 1, m jj = 0 ad m ij m ji. So, if m ji = 1 we have idetity i both cases ad if m ji = 0 we have complemetarity i both cases. Suppose all the rows of M are idetical. The, the first c + 1 colums are all-1 ad the rest of the colums are all-0. So, ay g.d has exactly c + 1 1s. So, a = b = c + 1, which is obviously ot the case. Thus, by Claim 6, we ca permute the rows ad colums to obtai a matrix M cosists of four submatrices M 1, M 2, M 3 ad M 4 of positive dimesios, where M 1 ad M 4 are all-1, ad M 2 ad M 3 are all-0 (Figure 9). Thus, F is rigid (Defiitio 5.2), cotrary to the hypothesis. We coclude that there must be a c-g.d i M. 14

15 Figure 9 I the case that the partitio E(G) = F (E(G)\F ) is rigid, if there exists a partitio P c+1 such that P c+1 F = c + 1, the clearly there is o partitio P c such that P c F = c. The proof of Theorem 5.6 shows that i this case, for ay c betwee a ad b there is a partitio P c such that 0 P c F c 1. Corollary 5.7. Let G = K, ad assume the partitio E(G) = F (E(G) \ F ) is ot rigid. The, there exist perfect matchigs P 1 ad P 2 such that P 1 F = ad P 2 F =. 6 Fair represetatio by perfect matchigs i K,, the case of three parts F 6.1 Statemet of the theorem ad outlie of proof I this sectio we prove Cojecture 1.11 for m = 3, amely: Theorem 6.1. Suppose that the edges of K, are partitioed ito sets E 1, E 2, E 3. The, there exists a perfect matchig F i K, satisfyig + 1 F E i 1 for every i = 1, 2, 3. Ei We idetify perfect matchigs i K, with permutatios i S. For σ, τ S, the Hammig distace (or plaily distace) d(σ, τ) is {i σ(i) τ(i)}. We write σ τ if d(σ, τ) 3. Let C be the simplicial complex of the cliques of this relatio. The proof will show, i effect, that η(c) 3. This will eable us to use Sperer s lemma. Here is a short outlie of this proof. Clearly, for each i 3 there exits a matchig F i represetig E i fairly, amely F i E i Ei Ei. We shall coect every pair F i, F j by a path cosistig of perfect matchigs represetig fairly E i E j, i which every two adjacet matchigs are -related. This forms a triagle D, which is ot ecessarily simple, amely it may have repeatig vertices, a triagulatio T of its circumferece, ad a assigmet A of matchigs to its vertices. We shall the show that there exists a triagulatio T extedig T ad cotaied i C. Here cotaied i C meas that there is a assigmet A of perfect matchigs to the vertices of T, that exteds A, such that the perfect matchigs assiged to adjacet vertices are -related. We color a vertex v of T by color i if A (v) represets 1 -fairly the set E i. By our costructio, this colorig satisfies the coditios of Sperer s lemma, ad applyig the lemma we obtai a multicolored triagle. We shall the show that at least oe of the matchigs assiged to the vertices of this triagle satisfies the coditio required i the theorem. F 15

16 6.2 A hexagoal versio of Sperer s Lemma I order to prove theorem 6.1 we shall use the followig versio of Sperer s lemma. Lemma 6.2. Let T be a piecewise liear image of the disc B 2, ad assume that its boudary is a (ot ecessarily simple) hexago H, with edges e 1,..., e 6. Suppose that the vertices of H are colored 1, 2, 3. Assume that No vertex i the triagulatio of e 1 has color 1. No edge i the triagulatio of e 2 is betwee two vertices of colors 1 ad 2. No vertex i the triagulatio of e 3 has color 2. No edge i the triagulatio of e 4 is betwee two vertices of colors 2 ad 3. No vertex i the triagulatio of e 5 has color 3. No edge i the triagulatio of e 6 is betwee two vertices of colors 3 ad 1. The T cotais a triagle betwee three vertices colored 1, 2 ad 3. Proof. We add three vertices to T utside the circumferece of the hexago so that a complex with a triagular boudary is formed: a vertex of color 1 adjacet to all vertices i the triagulatio of e 4, a vertex of color 2 adjacet to all vertices i the triagulatio of e 6 ad a vertex of color 3 adjacet to all vertices i the triagulatio of e 2. We ca the use the origial Sperer s Lemma o this augmeted complex. 6.3 Movig betwee permutatios Let i [] := {1,..., }. We defie a fuctio shift i : S S as follows: For every σ S, if σ(i) = j the for every k [] we defie i if k = i shift i (σ)(k) = j if σ(k) = i σ(k) otherwise Remark 6.3. Note that if σ(i) = i the shift i (σ) = σ. The aim of this operatio is to elimiate i i its cycles expressio, meaig obtaiig a permutatio sedig i to itself. The applicatio of shift i results i this state: if i is followed by j i some cycle of σ, the i is removed from this cycle. Lemma 6.4. Let be a positive umber, Let i [] ad let σ, τ S. If σ τ the shift i (σ) shift i (τ). Proof. For simplicity we write shift for shift i. Without loss of geerality i = 1. If shift(σ) = σ ad shift(τ) = τ the we are doe. The secod case we eed to cosider is that shift(τ) = τ but shift(σ) σ. Without loss of geerality τ = I. For every k [], if σ(k) = k the also shift(σ)(k) = k ad thus the distace betwee shift(σ) ad I is at most the distace betwee σ ad I, yieldig shift(σ) I = shift(τ). We are left with the case where shift(σ) σ ad shift(τ) τ. Without loss of geerality τ = (12) ad hece shift(τ) = I. As i the previous case, for every k [] if σ(k) = k the also shift(σ)(k) = k. We also ote that shift(σ)(1) = 1 but σ(1) 1 (sice shift(σ) σ). Therefore, the distace betwee shift(σ) ad I is strictly less tha the distace betwee σ ad I. If the distace betwee σ ad I is at 16

17 most 4 the shift(σ) I = shift(τ) ad we are doe. Sice σ τ, this distace caot be more tha 5, so it must be exactly 5. Note that if σ(1) = j 2, the σ ad τ differ o 1,2 ad j, ad thus σ(k) = k for all k {1, 2, j}, so the distace betwee σ ad I is at most 3, cotrary to the assumptio that this distace is 5. Thus, we must have that σ(1) = 2. It follows that the set A = {i [] : σ(i) τ(i)} is a set of size 3 disjoit from {1, 2}. But the also {i [] : shift(σ(i)) shift(τ(i))} = A, yieldig shift(σ) shift(τ). 6.4 Path coectivity Lemma 6.5. Let A = (a ij ) be a 0-1 matrix ad let k [ 1]. Let G be the graph whose vertices are the permutatios σ S satisfyig i=1 a iσ(i) k ad whose edges correspod to the relatio. If there exists ρ S with i=1 a iρ(i) > k, the G is coected. Proof. Without loss of geerality ρ is the idetity permutatio, amely i=1 a ii > k. We shall show that there is a path i G from ρ to σ for ay σ V (G) \ {ρ}. We prove this claim by iductio o d(σ, ρ). Write l = i=1 a iσ(i). Our aim is to fid distict j [] for which σ(j) j ad σ = shift j (σ) V (G). The the iductio hypothesis ca be applied sice σ σ ad σ is closer to ρ tha σ. If l k + 2 choose ay j [] with σ(j) j. The we have i=1 a iσ (i) i=1 a iσ(i) 2 k, so σ V (G). Suppose ext that l = k+1. By the assumptio that i=1 a ii > k we have i=1 a iσ(i) i=1 a ii ad sice σ ρ there must be some j [] for which σ(j) j ad a jj a jσ(j). Takig σ = shift j (σ) V (G) yields i=1 a iσ (i) i=1 a iσ(i) 1 = k, so σ V (G). Fially, if l = k the i=1 a iσ(i) < i=1 a ii ad hece there must be some j [] for which a jj > a jσ(j). Takig σ = shift j (σ) V (G) we get i=1 a iσ (i) i=1 a iσ(i) = k, so σ V (G). Corollary 6.6. Let A = (a ij ) be a 0-1 matrix ad let k []. Let G be the graph whose vertices are the permutatios σ S with i=1 a iσ(i) k ad whose edges correspod to the relatio. If i,j a ij k the G is coected. Proof. If there exists a permutatio ρ with i=1 a iρ(i) > k the we are doe by Lemma 6.5. If ot, by Köig s theorem there exist sets A, B [] with A + B k such that a ij = 0 for i A ad j B. This is compatible with the coditio i,j a ij k oly if A = 0 ad B = k or B = 0 ad A = k, ad a ij = 1 for all (i, j) A [] [] B. I both cases V (G) = S, implyig that the relatio is path coected sice every permutatio is reachable from every other permutatio by a sequece of traspositios. 6.5 Reachig a super-fair ad uder-fair represetatio simultaeously Lemma 6.7. Let the set E of edges of K, be partitioed to three sets E = E 1 E 2 E 3. The there exists a perfect matchig M with at least E1 edges of E 1 ad at most E3 edges of E 3. Proof. Let H be the graph with the edge set E 1 E 2. Köig s edge colorig theorem, combied with a easy alteratig paths argumet, yields that H ca be edge colored with colors i a way that each color class is of size either or. Clearly, at least oe of these classes cotais at least E 1 E(H) E(H) edges from E 1. A matchig with the desired property ca be obtaied by completig this color class i ay way we please to a perfect matchig of K,. We cojecture that a stroger property holds: 17

18 Cojecture 6.8. Let G = (V, E) be a bipartite graph with maximal degree ad let f : E {1, 2, 3,..., k} for some positive iteger k. The there exists a matchig M i G such that every umber j {1, 2, 3,..., k} satisfies {e E : f(e) j} {e M f(e) j} Clearly, we oly eed to see to it that the coditio holds for j < k. I [15], this cojecture was proved for G = K 6, A Sperer colorig of the boudary of a hexago From this poit util the ed of this sectio, we assume that the set E of edges of K, is partitioed to three sets E = E 1 E 2 E 3 ad for each i = 1, 2, 3 we have E i = k i where k i is iteger. We assume that Theorem 6.1 does ot hold, so i particular, there is o perfect matchig with exactly k i edges from each E i. As metioed i the previous sectio, Theorem 6.1 is easy i the case oe of the sets E i is empty, so we may assume k 1, k 2, k 3 {1,..., 2}. A perfect matchig is said to have property i+ if it has at least k i edges from E i, it is said to have property i + + if it has strictly more tha k i edges from E i ad it is said to have property i if it has at most k i edges from E i Figure 10 Lemma 6.9. There exists a triagulatio of the boudary of a hexago, ad a assigmet of a perfect matchig M v ad a color i v {1, 2, 3} to each vertex v of the triagulatio, such that M v has property i v + + ad the colorig satisfies the coditios of Lemma 6.2. Proof. By Lemma 6.7 there exists a perfect matchig M with properties 1+ ad 3. We assig it to oe vertex of the hexago. By permutig the roles of E 1, E 2, E 3 we ca fid six such perfect matchigs ad assig them to the six vertices of the hexago as i Figure 10. By Corollary 6.6, we ca fill the path betwee the two permutatios with property i i a way that all perfect matchigs i the path have property i. Similarly, we ca fill the path betwee the two permutatios with property i+. For each vertex v we assig a color i v such that M v has property i v + +. Now suppose Lemma 6.2 does ot hold, the without loss of geerality we have two perfect matchigs 18

19 M 1 M 2, where M 1 has properties 3+ ad ad M 2 has properties 3+ ad ad this easily gives Theorem Simple coectivity I the ext two lemmas let be a positive iteger, i [] ad σ.τ S. We write shift for shif i. Lemma If d(σ, τ) = 2, the the 4-cycle σ τ shift(τ) shift(σ) σ is a ull cycle i C (i.e., it ca be triagulated.) Proof. If either σ shift(τ) or τ shift(σ) the we are doe. So, we may assume this does ot happe ad i particular σ shift(σ) ad τ shift(τ). We may assume, without loss of geerality, that i = 1, σ = (12), τ = (12)(34), shift(σ) = I ad shift(τ) = (34). We ca ow fill the cycle as i Figure 11. Figure 11 Lemma If d(σ, τ) = 3 the the 4-cycle σ τ shift(τ) shift(σ) σ is a ull cycle i C. Proof. Let ρ S have distace 2 from both σ ad τ. Deote σ = shift(σ), τ = shift(τ) ad ρ = shift(ρ). We use the previous lemma to fill the cycle as i Figure 12. Lemma The simplicial complex C is simply coected. Proof. Let f : S 1 C be a cotiuous fuctio ad let Z = Imf. We eed to show that Z is a ull cycle i C by fidig a cotiuous fuctio g : B 2 C extedig f. We arbitrarily pick distict i [], ad wheever a poit p S 1 has f(p) = σ Z, we defie g(p/2) = σ = shift i (σ). We ow use the previous two lemmas to fill the outer shell of B 2. (See Figure 13.) We repeat this for all i ad evetually, we are left with a cycle all of whose poits are set to the idetity permutatio. We defie g to be I i the iterior of this cycle as well ad we are doe. 19

20 Figure 12 Figure Applyig Sperer s lemma We form a matrix A = (a ij ) i,j, where a ij = p (p = 1, 2, 3) if the edge ij belogs to E p. For each l {1, 2, 3} ad σ S we write d l (σ) = {i : a iσ(i) = l} k l. Our ext lemma poits at what we may aspire for, i order to prove the theorem. Lemma Suppose that the triple {σ 1, σ 2, σ 3 } is i C, ad that d l (σ l ) > 0 for each l {1, 2, 3}. The there exists σ S with d l (σ) 1 for each l {1, 2, 3}. Sice the existece of such σ 1, σ 2, σ 3 follows from Lemmas 6.2, 6.9 ad 6.12, this will fiish the proof 20

21 of Theorem 6.1. Proof. Defie a 3 3 matrix B = (b ij ) by b ij = d i (σ j ). We kow that the diagoal etries i B are positive, the sum i each colum is zero, ad ay two etries i the same row differ by at most 3. This meas that the miimal possible etry i B is -2. We may assume each colum has some etry ot i { 1, 0, 1}. Let us start with the case that all of the diagoal etries of B are at least 2. This implies that all off-diagoal etries are at least -1. Sice each colum must sum up to zero, we must have B = This implie that the distace betwee ay two of σ 1, σ 2, σ 3 is exactly 3, ad without loss of geerality σ 1 = I, σ 2 = (123), σ 3 = (132), ad the matrix A has the form A = We ca ow take σ = (12) ad we are doe. We are left with the case that some diagoal etry of B is 1. Without loss of geerality b 11 = 1. We also assume without loss of geerality that b 21 b 31. Sice the first colum must sum up to zero, we have b 21 + b 31 = 1, ad thus 0.5 = 0.5(b 21 + b 31 ) b 31 = 1 b I other words, either b 21 = 1 ad b 31 = 0 or b 21 = 2 ad b 31 = 1. I the first case we ca just take σ = σ 1 ad we are doe. Therefore we assume the secod case. B = Sice d 3 (σ 1 ) > 0, we may assume σ 3 = σ 1, ad due to the -2 etries i the secod row, we must have b 22 = 1. We ow get 1 1 B = Without loss of geerality b 12 b 32 ad by argumets similar to the above we ca fill the secod colum B = The distace betwee σ 1 ad σ 2 is exactly 3, so without loss of geerality σ 1 = I ad σ 2 = (123). I order to achive the values of b 12 = 2, b 11 = 1, b 21 = 2, b 22 = 1 we must have a ii = 1 ad a iσ2(i) = 2 for each i {1, 2, 3}. The oly case i which oe of the choices σ = (12) or σ = (23) or σ = (13) works is if a 13 = a 21 = a 32 = 3, so oce agai we get 21

22 A = We have b 31 = 1 which meas that 3 appears k times o the diagoal. Without loss of geerality a 44 = a 55 =... = a k3+4 k 3+4 = 3. I ay of the followig cases oe ca easily fid some σ S with d l (σ) 1 for each l {1, 2, 3}: If either a ij 3 or a ji 3 for some i {4,..., k 3 + 4} ad j {1, 2, 3}. If a ij 3 for some i, j {4,..., k 3 + 4} If both a ij 3 ad a ji 3 for some i {4,..., k 3 + 4} ad j {k 3 + 5,..., }. If oe of the above occures the which is a cotradictio. k 3 = {(i, j) : a ij = 3} 2 3 (1 + k 3 ) + (1 + k 3 ) (k 3 + 1)( k 3 4) Remark After the articulatio of the above topological proof of Theorem 6.1, a combiatorial proof was give i [16]. Refereces [1] M. Adamaszek ad J. A. Barmak, O a lower boud for the coectivity of the idepedece complex of a graph, Discrete Math. 311 (2011), [2] R. Aharoi, R. Holzma, D. Howard ad P. Sprüsell, Cooperative colorigs ad systems of idepedet represetatives, submitted for publicatio [3] R. Aharoi, N. Alo ad E. Berger, Eigevalues of K 1,k -free graphs ad the coectivity of their idepedece complexes submitted for publicatio [4] R. Aharoi, J. Barat ad I. Waless, Raibow matchig submitted for publicatio [5] R. Aharoi, E. Berger ad R. Ziv, Idepedet systems of represetatives i weighted graphs, Combiatorica 27 (2007), [6] R. Aharoi ad P. Haxell, Hall s theorem for hypergraphs, J. of Graph Theory 35 (2000), [7] R. Aharoi, I. Gorelik ad L. Naris, Coectivity of the idpedece complex of lie graphs, i preparatio [8] R. Aharoi, R. Maber ad B. Wajryb, Special parity of perfect matchigs i bipartite graphs, Discrete Math. 79 (1989/1990), [9] N. Alo, Splittig ecklaces, Advaces i Mathematics 63 (1987), [10] N. Alo, The liear arboricity of graphs, Israel J. Math., 62 (1988),